Class Solution
Hard
You are given two strings s
and t
of equal length n
. You can perform the following operation on the string s
:
- Remove a suffix of
s
of lengthl
where0 < l < n
and append it at the start ofs
.
For example, lets = 'abcd'
then in one operation you can remove the suffix'cd'
and append it in front ofs
makings = 'cdab'
.
You are also given an integer k
. Return the number of ways in which s
can be transformed into t
in exactly k
operations.
Since the answer can be large, return it modulo 109 + 7
.
Example 1:
Input: s = “abcd”, t = “cdab”, k = 2
Output: 2
Explanation:
First way:
In first operation, choose suffix from index = 3, so resulting s = “dabc”.
In second operation, choose suffix from index = 3, so resulting s = “cdab”.
Second way:
In first operation, choose suffix from index = 1, so resulting s = “bcda”.
In second operation, choose suffix from index = 1, so resulting s = “cdab”.
Example 2:
Input: s = “ababab”, t = “ababab”, k = 1
Output: 2
Explanation:
First way:
Choose suffix from index = 2, so resulting s = “ababab”.
Second way:
Choose suffix from index = 4, so resulting s = “ababab”.
Constraints:
2 <= s.length <= 5 * 105
1 <= k <= 1015
s.length == t.length
s
andt
consist of only lowercase English alphabets.
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Constructor Summary
Constructors -
Method Summary
Modifier and TypeMethodDescriptionint
numberOfWays
(java.lang.String s, java.lang.String t, long k) Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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numberOfWays
public int numberOfWays(java.lang.String s, java.lang.String t, long k)
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